In two concentric circles centred at $O$, a chord $AB$ of the larger circle touches the smaller circle at $C$. If $OA = 3.5$ cm, $OC = 2.1$ cm, then $AB$ is equal to

In two concentric circles centred at $O$, a chord $AB$ of the larger circle touches the smaller circle at $C$. If $OA = 3.5$ cm, $OC = 2.1$ cm, then $AB$ is equal to:

We are given that $AB$ is a chord of the larger circle that touches the smaller circle at $C$. Since $AB$ touches the smaller circle at $C$, the radius $OC$ is perpendicular to $AB$. This means $OC$ bisects $AB$ at $C$.

Let $AB = 2x$, so that $AC = x$. In right triangle $OCA$, using the Pythagorean theorem:

$OA^2 = OC^2 + AC^2$

Substituting the given values:

$(3.5)^2 = (2.1)^2 + x^2$

$12.25 = 4.41 + x^2$

$x^2 = 12.25 - 4.41 = 7.84$

$x = \sqrt{7.84} = 2.8$

Since $AB = 2x$, we get:

$AB = 2(2.8) = 5.6$ cm

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